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Question

Question: Define Equal Function....

Define Equal Function.

Explanation

Solution

We will use the concept and definition of function as applicable to the set theory. Equal function is one in which both Domain and Range will be the same.

Complete step by step solution: Let A and B be two sets. Now by function we mean a rule such that for aA\forall a \in Athere exists a unique bBb \in Bsuch that, f(a)=bf(a) = b.SO, the set A is called the Domain of ffwile the set B is called the Range offf. Now, let f:ABf:A \to Band ggbe another function such that g:ABg:A \to B. So, the functions ffand ggwill be equal functions if and only if their Domain and Range are the same.
That is, we can say that, if for the function f:ABf:A \to B and g:ABg:A \to B ,aA\forall a \in Athere exists bBb \in B, such that f(a)=g(a)f(a) = g(a)then the two functions ffand ggare equal functions.
For example,
Let us take two functions f(x)=1xf\left( x \right) = \dfrac{1}{x}and g(x)=xx2g\left( x \right) = \dfrac{x}{{{x^2}}}. Now the function g(x)=xx2g\left( x \right) = \dfrac{x}{{{x^2}}}, can be simplified and reduced to g(x)=1xg\left( x \right) = \dfrac{1}{x}. Now the domain of f(x)f(x)will be all real numbers except 0 and the domain of the reciprocal function g(x)g(x)will also be all real numbers except 0.
Therefore, the domain of both the functions is the same.
Similarly, since both the functions will take all values across the Real numbers, so their Range will also be the same. Such that f(x)=g(x)f(x) = g(x).
Thus both the functions f(x)f(x)and g(x)g(x) will be equal functions.

Note: Apart from equal functions there are other types of functions also. Let us consider two sets A and B. A one-to-one function is one where every element belonging to a set A, that has exactly one unique image in set B. An onto function is one where every element belonging to set B has at least one preimage in the set A. A function which is both one-to-one and onto, is called an injective function. If either of the characteristics is missing, the function will not be an injective function.